Statics, fourteenth edition by r c hibbeler section 8 2

In-Class Activities:• Check Homework, if any Students will be able to: a Determine the forces on a wedge.. How can we determine the force required to pull the wedge out?When there are no

In-Class Activities:

• Check Homework, if any

Students will be able to:

a) Determine the forces on a wedge

b) Determine tension in a belt

WEDGES AND FRICTIONAL FORCES ON FLAT BELTS

1 A wedge allows a force P to lift

a _ weight W

A) (large, large) B) (small, large)

C) (small, small) D) (large, small)

2 Considering friction forces and the

indicated motion of the belt, how are belt

tensions T1 and T2 related?

A) T1 > T2 B) T1 = T2

C) T1 < T2 D) T1 = T2 e

READING QUIZ

How can we determine the force required to pull the wedge out?

When there are no applied forces on the wedge, will it

stay in place (i.e., be self-locking) or will it come out on its own? Under what physical conditions will it come

out?

Wedges are used to adjust the elevation or provide stability for heavy objects such as this large steel pipe

APPLICATIONS

How can we decide if the belts will function properly, i.e., without slipping or

breaking?

Belt drives are commonly used for transmitting the torque

developed by a motor to a wheel attached to a pump, fan

or blower

APPLICATIONS (continued)

How can you determine the tension in the cable pulling on the band?

Also from a design perspective, how are the belt tension, the applied force P and the torque M, related?

In the design of a band brake, it

is essential to analyze the frictional forces acting on the band (which acts like a belt)

APPLICATIONS (continued)

It is easier to start with a FBD of the wedge since you know the direction of its motion.

Note that:

a) the friction forces are always in the

direction opposite to the motion , or impending motion, of the wedge;

b) the friction forces are along the contacting surfaces; and,

c) the normal forces are perpendicular to the contacting surfaces.

To determine the unknowns, we must apply E-of-E,  Fx = 0 and

 Fy = 0, to the wedge and the object as well as the impending motion frictional equation, F = S N

Next, a FBD of the object on top of the wedge

is drawn Please note that: a) at the contacting surfaces between the

wedge and the object, the forces are equal in magnitude and opposite in direction to those

on the wedge; and, b) all other forces acting on the object should

be shown

ANALYSIS OF A WEDGE (continued)

Now of the two FBDs, which one should we start

analyzing first?

We should start analyzing the FBD in which the number

of unknowns are less than or equal to the number of

E-of-E and frictional equations

ANALYSIS OF A WEDGE (continued)

If the object is to be lowered, then the wedge needs to be pulled out If the value

of the force P needed to remove the wedge

is positive, then the wedge is self-locking, i.e., it will not come out on its own

ANALYSIS OF A WEDGE (continued)

Detailed analysis (please refer to your textbook) shows that

T2 = T1 e   where  is the coefficient of static friction

between the belt and the surface Be sure to use radians

when using this formula!!

If the belt slips or is just about to slip,

then T2 must be larger than T1 and the motion resisting friction forces Hence,

T2 must be greater than T1

Consider a flat belt passing over a fixed curved surface with the total angle of contact equal to  radians

BELT ANALYSIS

1 Draw FBDs of wedge A and wedge B

2 Apply the E-of-E to wedge B Why do wedge B first?

3 Apply the E-of-E to wedge A

EXAMPLE

Given: The 3000-lb load is applied to

wedge B The coefficient of static friction between A and C and between B and D is 0.3, and between A and B it is 0.4

Assume the wedges have negligible weight

Find: The smallest force P needed to lift

3000 lb load

Plan:

Applying the E-of-E to wedge B, we get

Applying the E-of-E to wedge A, we get

2 The boy (hanging) in the picture weighs

100 lb and the woman weighs 150 lb The

coefficient of static friction between her

shoes and the ground is 0.6 The boy will

?

A) Be lifted up B) Slide down

C) Not be lifted up D) Not slide down

1 Determine the direction of the friction

force on object B at the contact point

between A and B

A) B) 

C)  D)

CONCEPT QUIZ

Given: A force P is applied to move

wedge A to the right The spring is compressed a distance of 175 mm

The static friction coefficient

is S = 0.35 for all contacting surfaces Neglect the weight

1 Draw FBDs of block B and wedge A

2 Apply the E-of-E to block B to find the friction force when the wedge is on the verge of moving

3 Apply the E-of-E to wedge A to find the smallest force

needed to cause sliding

GROUP PROBLEM SOLVING (continued)

If the wedge is on the verge of moving to the right, then slipping will have to occur at both contact surfaces

Thus, FA = S NA = 0.35 NA and FB = 0.35 NB

GROUP PROBLEM SOLVING (continued)

Using the spring formula:

Fsp = K x = (15 kN/m) (0.175m) = 2.625 kN

NC

FSP = 15(0.175) = 2.625 kN

+  FY = NB – 2.625 = 0

NB = 2.625 kN

Applying the E-of-E to the Block B, we get:

GROUP PROBLEM SOLVING (continued)

+  FY = NA cos 10 – 0.35NA sin 10 – 2.625 = 0

NA = 2.841 kN

+  FX = P – 0.35(2.625) –0.35(2.841) cos10– 2.841 sin10 = 0 Applying the E-of-E to Wedge A:

2 In the analysis of frictional forces on a flat belt, T2 = T1 e  

In this equation,  equals

A) Angle of contact in degrees B) Angle of contact in radians C) Coefficient of static friction D) Coefficient of kinetic friction

1 When determining the force P needed to lift the

block of weight W, it is easier to draw a FBD

of first

A) The wedge B) The block

C) The horizontal ground D) The vertical wall

W

ATTENTION QUIZ

End of the Lecture

Let Learning Continue